A 2.05-cm-tall object is placed 30.0 cm to the left of a converging lens with a

Question

A 2.05-cm-tall object is placed 30.0 cm to the left of a converging lens with a focal length f1=20.5 cm. A diverging lens, with a focal length f2=-42.5 cm, is placed 30.0 cm to the right of the first lens. How tall is the final image of the object?

I calculated the image distance from lens 1 to be +64.74 cm which means it is to the right of both the first and second lens. Then, the next step was doing 30 cm - 64.74 cm = -34.74 cm which is the object distance from lens 2. Then I found the image distance from lens 2 to be +190.26cm. These are the correct answers. What I don't understand is:

1) Why we didn't do 64.74cm - 30cm to find the object distance from lens 2?

2) How can the object distance be a negative value? (-34.74cm)

3) How does a negative value object distance result in a positive value image distance when the object and image end up being on the same side of the lens?  

Thanks!

Explanation / Answer

Ans

1) The object distance is the distance of the object from the lens. Distance is a positive number so we get the same number from 64.74 - 30 and |30 - 64.74|(absolute value). So it won't matter which way you find the distance.

2) Now we apply sign conventions to get the object distance(so far we have the distance only) Outgoing light from the two lenses is to the right, and if an object is on the same side as outgoing light, then it is negative number i.e distance with negative sign in front of it.

3) Notice the second lens is diverging lens. So it's image has to be on the right side(object and image are on the same side for a divergeing lens) Since image is on the same side as outgoing light ie to right of the 2nd lens(lenses) so it is real and real images have positive distances.

It has to do with sign conventions.

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